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NeTakaya
3 years ago
12

A 70.0-kg person throws a 0.0420-kg snowball forward with a ground speed of 35.0 m/s. A second person, with a mass of 57.0 kg, c

atches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.20 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged?
Physics
1 answer:
ollegr [7]3 years ago
3 0

Answer:

so throwers new velocity = 2.18032m/s

so catchers new velocity = 0.02577m/s

Explanation:

Directly by conservation of momentum we can write

m_1u_1+m_2u_2= m_1v_1+m_2v_2

let x be the thrower's new velocity

(70+0.042)×2.2 + 57×0 = 70× x +0.042×35 +57×0

x = 2.18032m/s

so the velocity of 70 kg man = 2.18032m/s

so throwers new velocity = 2.18032m/s

now again by conservation of momentum

0.042×35 = (57+0.042) ×y

y = 0.02577m/s

so catchers new velocity = 0.02577m/s

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A runner starts from rest and in 3 s reaches a speed of 8 m/s. If we assume that the speed changed at a constant rate (constant
Stells [14]

Answer:

The average speed of the runner is 4 m/s.

Explanation:

Hi there!

The average speed (a.s) is calculated by dividing the traveled distance (d) over the time needed to travel that distance (t):

a.s = d / t

So, let´s find the distance traveled in those 3 s. For that, we can use the equation of position of an object moving in a straight line with constant acceleration:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

If we place the origin of the frame of reference at the point where the runner starts, then, x0 = 0. Since the runner starts from rest, v0 = 0. So, the equation gets reduced to this:

x = 1/2 · a · t²

We have the time (3 s), so let´s find the acceleration. For that, we can use the equation of velocity of an object moving in a straight line with constant acceleration:

v = v0 + a · t

Where "v" is the velocity at a time "t". Since v0 = 0, then:

v = a · t

At t = 3 s, v = 8 m/s

8 m/s = a · 3 s

8/3 m/s² = a

So let´s find the position of the runner at t = 3 s (In this case, the position of the runner will be equal to the traveled distance):

x = 1/2 · a · t²

x = 1/2 · 8/3 m/s² · (3 s)²

x = 12 m

Now, we can calcualte the average speed:

a.s = d/t

a.s = 12 m / 3 s

a.s = 4 m/s

The average speed of the runner is 4 m/s.

4 0
3 years ago
6. Which of the following best describes the situation?
Alla [95]

Answer:

D

Explanation:

the man is pull the box that is the force and the box rubbing against the ground causes friction

7 0
3 years ago
What kind of charging occurs when you slide your body across a plastic surface?
Romashka [77]
<span>Charging by friction occurs, Electrons are transferred when one object rubs against another.

Another example of this would be socks on carpet.

Hope this helps!</span>
6 0
4 years ago
Four-wheel drive trucks do not stop better on icy
MaRussiya [10]
It would be Newton’s second law of motion
8 0
3 years ago
An object starts from rest and has an acceleration given by a = 3.00 t^2 – 4.20 t. Determine how far the object is from its star
Korolek [52]

Answer:

38.7 units

Explanation:

a = 3.00 t² – 4.20 t

Integrate to find velocity as a function of time.

v = ∫ a dt

v = ∫ (3.00 t² – 4.20 t) dt

v = 1.00 t³ – 2.10 t² + C

The object starts at rest, so at t = 0, v = 0.

0 = 1.00 (0)³ – 2.10 (0)² + C

0 = C

v = 1.00 t³ – 2.10 t²

Integrate to get position as a function of time.

x = ∫ v dt

x = ∫ (1.00 t³ – 2.10 t²) dt

x = 0.250 t⁴ – 0.700 t³ + C

Find the difference in positions between t = 4.50 and t = 0.

Δx = [0.250 (4.50)⁴ – 0.700 (4.50)³ + C] – [0.250 (0)⁴ – 0.700 (0)³ + C]

Δx = 0.250 (4.50)⁴ – 0.700 (4.50)³

Δx = 38.7

The object moves 38.7 units.

7 0
3 years ago
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